Problem: Given the equation: $ y = -3x^2 + 6x - 4$ Find the parabola's vertex.
Answer: When the equation is rewritten in vertex form like this, the vertex is the point $({h}, {k})$ $ y = A(x - {h})^2 + {k} $ We can rewrite the equation in vertex form by completing the square. First, move the constant term to the left side of the equation: $ \begin{eqnarray} y &=& -3x^2 + 6x - 4 \\ \\ y + 4 &=& -3x^2 + 6x \end{eqnarray} $ Next, we can factor out a $-3$ from the right side: $ y + 4 = -3(x^2 - 2x) $ We can complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $-2$ , so half of it would be $-1$ , and squaring that gives us ${1}$ . Because we're adding the $1$ inside the parentheses on the right where it's being multiplied by $-3$ , we need to add ${-3}$ to the left side to make sure we're adding the same thing to both sides. $ \begin{eqnarray} y + 4 &=& -3(x^2 - 2x) \\ \\ y + 4 + {-3} &=& -3(x^2 - 2x + {1}) \\ \\ y + 1 &=& -3(x^2 - 2x + 1) \end{eqnarray} $ Now we can rewrite the expression in parentheses as a squared term: $ y + 1 = -3(x - 1)^2 $ Move the constant term to the right side of the equation. Now the equation is in vertex form: $ y = -3(x - 1)^2 - 1 $ Now that the equation is written in vertex form, the vertex is the point $({h}, {k})$ $ y = A(x - {h})^2 + {k} $ $ y = -3(x - {(1)})^2 + {(-1)} $ The vertex is $({1}, {-1})$. Be sure to pay attention to the signs when interpreting an equation in vertex form.